![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: Các dấu bằng ở yêu cầu là dấu cộng.
1. Có: \(x+y=3\)
\(\Leftrightarrow\left(x+y\right)^2=3^2\)
\(\Leftrightarrow x^2+2xy+y^2=9\)
\(\Leftrightarrow x^2+y^2=9-2\cdot1=7\) (do \(xy=1\))
\(------\)
Lại có: \(x+y=3\)
\(\Leftrightarrow\left(x+y\right)^3=3^3\)
\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=27\)
\(\Leftrightarrow x^3+y^3+3\cdot1\cdot3=27\) (do x + y = 3; xy = 1)
\(\Leftrightarrow x^3+y^3=18\)
Ta có: \(x^2+y^2=7\)
\(\Leftrightarrow\left(x^2+y^2\right)^2=7^2\)
\(\Leftrightarrow x^4+y^4+2\cdot\left(xy\right)^2=49\)
\(\Leftrightarrow x^4+y^4=49-2\cdot1=47\) (do xy = 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhờ các bạn giúp. Mình cần gấp. Cảm ơn!
Bài 1; Cho biểu thức: B= (x2 +1)(y2 + 1) - (x+4)(x-4) - (y-5)(y+5)
a) CMR: B 42 với mọi giá trị của x và y
b) Tìm x và y để B= 42
Giải:
a) B = (x2 +1)(y2 + 1) - (x+4)(x-4) - (y-5)(y+5)
B = \(x^2y^2+x^2+y^2+1-x^2+16-y^2+25\)
B = \(x^2y^2+42\ge42\) với mọi x , y
b) Để B = 42 \(\Rightarrow\) x2y2 + 42 = 0 \(\Rightarrow\) x2y2 = 0 \(\Rightarrow\) x = y = 0
Bài 2:
a) Tìm GTNN của A= (x- 1)(x+ 2)(x+ 3)(x+6)
b) Tìm GTNN cuả B= 3xy(x+ 3y) - 2xy(x+4y) - x2(y-1) + y2(1-x) + 36
Giải:
a) A = (x-1)(x+2)(x+3)(x+6)
A = (x2 + 5x - 6)(x2 + 5x + 6)
A = ( x2 + 5x )2 - 36 \(\ge\) -36 với mọi x
Dấu " = " xảy ra khi x2 + 5x = 0
x ( x + 5 ) = 0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
MinA = -36 khi và chỉ khi x = 0 hoặc x = -5
b) Chịu :))
Bài 1:
a) \(\left(x^2+1\right)\left(y^2+1\right)-\left(x+4\right)\left(x-4\right)-\left(y-5\right)\left(y+5\right)\)
\(=x^2y^2+x^2+y^2+1-x^2+16-y^2+25\)
\(=x^2y^2+42\ge42\forall x\) (đpcm)
b) Để B = 42 thì \(x^2y^2+42=42\)
\(\Leftrightarrow x^2y^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) (1)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)
\(\Leftrightarrow26x+28=42\)
\(\Leftrightarrow26x=42-28\)
\(\Leftrightarrow26x=14\)
\(\Leftrightarrow x=\dfrac{7}{13}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{7}{13}\right\}\)
1b) \(5x\left(x+3\right)^2-5\left(x+1\right)^3+15\left(x+2\right)\left(x-2\right)=5\Leftrightarrow5x\left(x^2+6x+9\right)-5\left(x^3+3x^2+3x+1\right)+15\left(x^2-4\right)=5\Leftrightarrow30x-65=5\Leftrightarrow30x=70\Leftrightarrow x=\dfrac{7}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1+2\right),y^2-13y+12=y^2-12y-y-12=y\left(y-12\right)+\left(y-12\right)=\left(y+1\right)\left(y-12\right)\)
\(3,x^2-x-30=x^2-6x+5x-30=x\left(x-6\right)+5\left(x-6\right)=\left(x+5\right)\left(x-6\right)\)
\(4,y^2+y-42=y^2-6y+7y-42=y\left(y-6\right)+7\left(y-6\right)=\left(y+7\right)\left(y-6\right)\)
\(5,x^2+3x-10=x^2-2x+5x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
\(6,x^2-8x+15=x^2-5x-3x+15=x\left(x-5\right)-3\left(x-5\right)=\left(x-3\right)\left(x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2=3\frac{1}{3}xy\)hay \(x^2+y^2=\frac{10}{3}xy\)
\(\Rightarrow x^2+2xy+y^2=\frac{16}{3}xy\)\(\Rightarrow\left(x+y\right)^2=\frac{16}{3}xy\)
tương tự : \(\left(x-y\right)^2=\frac{4}{3}xy\)
\(\Rightarrow\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{1}{4}\Rightarrow\orbr{\begin{cases}\frac{x-y}{x+y}=\frac{1}{2}\\\frac{x-y}{x+y}=\frac{-1}{2}\end{cases}}\)
vì x > y > 0 nên x - y > 0 \(\Rightarrow\frac{x-y}{x+y}>0\)
Vậy \(\frac{x-y}{x+y}=\frac{1}{2}\)
Xét\(x^2+2xy+y^2=\frac{10}{3}xy+2xy=\frac{16}{3}xy\)
\(x^2-2xy+y^2=\frac{10}{3}xy-2xy=\frac{4}{3}xy\)
Từ đó ta được:
\(\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{\left(\frac{4}{3}xy\right)}{\left(\frac{16}{3}xy\right)}=\frac{1}{4}\)
\(\Rightarrow\sqrt{\frac{\left(x-y\right)^2}{\left(x+y\right)^2}}=\frac{1}{2}\Rightarrow\left|\frac{x-y}{x+y}\right|=\frac{1}{2}\)
Hihi
đến đây bạn tự làm nốt nha
^-^ Học tốt
\(y\left(y+1\right)\left(y^2+y+1\right)\Leftrightarrow\left[y\left(y+1\right)\right]\left[y\left(y+1\right)+1\right]=42\)
\(Đặt:y\left(y+1\right)=t.Taco:t\left(t+1\right)=42\Leftrightarrow t^2+t+\frac{1}{4}=\frac{169}{4}\Leftrightarrow\left(t+\frac{1}{2}\right)^2=\frac{169}{4}\Leftrightarrow\left[{}\begin{matrix}t+\frac{1}{2}=-\frac{13}{2}\\t+\frac{1}{2}=\frac{13}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-7\\t=6\end{matrix}\right.\)\(+,t=-7\Rightarrow y\left(y+1\right)=-7\Leftrightarrow y^2+y=-7\Leftrightarrow y^2+y+\frac{1}{4}=-\frac{27}{4}\Leftrightarrow\left(y+\frac{1}{2}\right)^2=-\frac{27}{4}\left(\text{ loại}\right)\)
\(+,t=6\Rightarrow y^2+y=6\Leftrightarrow y^2+y+\frac{1}{4}=\frac{25}{4}\Leftrightarrow\left(y+\frac{1}{2}\right)^2=\frac{25}{4}\Leftrightarrow\left[{}\begin{matrix}y+\frac{1}{2}=-\frac{5}{2}\\y+\frac{1}{2}=\frac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-3\\y=2\end{matrix}\right..Vậy:y\in\left\{-3;2\right\}\)